---
title: "U/B diagram, BCR with PW, discounted payback"
subject: "Engineering Economics"
chapter: 3
key: "semester_6/engineering_economics:ub-diagram-bcr-pw-dpp"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/ub-diagram-bcr-pw-dpp
site: BCT Engineering
---

# U/B diagram, BCR with PW, discounted payback

Engineering Economics · Chapter 3: Basic Methodologies of Engineering Economic Analysis

- Asked in: 2070 Bhadra, 6+5+5 marks
- Video: [#ECO_4.3 Both type of BC ratio using PW and AW | 016 spring 3b Engineering Economics - Shiva Gyawali](https://www.youtube.com/watch?v=AOsUV9eWHtY) — Shiva Gyawali
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/ub-diagram-bcr-pw-dpp
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-3

## Question

Initial Investment = Rs. 100,000; Salvage Value = 0; Annual O&M Cost = Rs. 20,000; Useful Life = 5 years; Annual Benefit = 60,000 at the end of first year, thereafter decreases by 4,000 each year for the remaining years.
a) Draw U/B diagram.
b) Evaluate conventional BCR using PW formulation. Take salvage value = 10,000.
c) Evaluate Discounted Payback Period. Take standard (cut off) Payback Period = 3 years.

## Short answer

Unrecovered balance becomes positive in year 4. Conventional B/C = 200,000/(100,000+75,816−6,209)=1.18 (acceptable). Discounted payback = 3.51 years > 3 years, so project fails payback test.

## Answer

The MARR is not given; assumption: MARR = 10% per year.

#### Net cash flow
| Year | Benefit | O&M | Net CF |
|---|---|---|---|
| 0 | | | −1,00,000 |
| 1 | 60,000 | 20,000 | 40,000 |
| 2 | 56,000 | 20,000 | 36,000 |
| 3 | 52,000 | 20,000 | 32,000 |
| 4 | 48,000 | 20,000 | 28,000 |
| 5 | 44,000 | 20,000 | 24,000 |

#### a) Unrecovered balance (U/B) diagram (salvage = 0, i = 10%)
B_t = B_(t−1)(1.10) + net CF_t

| Year | B_(t−1) × 1.10 | Net CF | B_t |
|---|---|---|---|
| 1 | -110,000 | 40,000 | -70,000 |
| 2 | -77,000 | 36,000 | -41,000 |
| 3 | -45,100 | 32,000 | -13,100 |
| 4 | -14,410 | 28,000 | 13,590 |
| 5 | 14,949 | 24,000 | 38,949 |

```text
 Unrecovered balance (each # = about 4,000)
 yr0:  -100,000 |-#########################
 yr1:   -70,000 |-##################
 yr2:   -41,000 |-##########
 yr3:   -13,100 |-###
 yr4:    13,590 |+###
 yr5:    38,949 |+##########
```
The balance is negative until between years 3 and 4, where it crosses zero (about 3.5 years), and ends at +38,949 at year 5, which is the future worth of the project's surplus.

#### b) Conventional BCR using PW (S = 10,000)
- PW(B) = 60,000 (P/F,10%,1) + 56,000 (P/F,10%,2) + 52,000 (P/F,10%,3) + 48,000 (P/F,10%,4) + 44,000 (P/F,10%,5) = Rs 200,000
  (check: 60,000 (P/A,10%,5) − 4,000 (P/G,10%,5) = 227,447 − 27,447)
- PW(O&M) = 20,000 × (P/A, 10%, 5) = 20,000 × 3.7908 = Rs 75,816
- PW(S) = 10,000 × (P/F, 10%, 5) = 10,000 × 0.6209 = Rs 6,209

B/C = PW(B) / [I + PW(O&M) − PW(S)] = 200,000 / (1,00,000 + 75,816 − 6,209) = 200,000 / 169,607 = **1.179**

Since B/C > 1, the project is acceptable.

#### c) Discounted payback period (cut-off = 3 years, S = 0)
| Yr | Benefit | Net CF | (P/F,10%,t) | PW of net CF | Cumulative PW − 1,00,000 |
|---|---|---|---|---|---|
| 1 | 60,000 | 40,000 | 0.9091 | 36,364 | -63,636 |
| 2 | 56,000 | 36,000 | 0.8264 | 29,752 | -33,884 |
| 3 | 52,000 | 32,000 | 0.7513 | 24,042 | -9,842 |
| 4 | 48,000 | 28,000 | 0.6830 | 19,124 | 9,282 |
| 5 | 44,000 | 24,000 | 0.6209 | 14,902 | 24,184 |

After 3 years, the unrecovered amount is 9,842. Year 4 recovers PW = 19,124.

θ_d = 3 + 9,842 / 19,124 = **3.51 years**

**Answer: B/C ≈ 1.18 (> 1, accept); discounted payback = 3.51 years > 3 years, so the project fails the discounted payback test.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is yearly net cash flow found?
   A. Benefit plus O&M
   B. Benefit minus O&M
   C. Benefit only
   D. O&M minus benefit
   Answer: B. E.g. year 1: 60,000 - 20,000 = 40,000. (Key point: Net CF)

2. How is the unrecovered balance computed?
   A. B_t = B_(t-1)(1.10) + net CF_t
   B. B_t = B_(t-1) + net CF_t
   C. B_t = B_(t-1)/1.10
   D. B_t = net CF_t x 1.10
   Answer: A. The balance earns interest at MARR and receives each net flow. (Key point: U/B recursion)

3. Which expression is the conventional BCR?
   A. PW(B)/I
   B. [PW(B) - PW(O&M)]/I
   C. PW(B) / [I + PW(O&M) - PW(S)] = 1.179
   D. PW(B) / PW(O&M)
   Answer: C. 200,000 / (100,000 + 75,816 - 6,209); greater than 1 so acceptable. (Key point: Conventional BCR)

4. What is the discounted payback and verdict?
   A. 2.51 years, passes
   B. 3.51 years, passes
   C. 3.00 years, passes
   D. 3.51 years, fails the 3-year cut-off
   Answer: D. Unrecovered 9,842 after year 3 divided by year-4 PW 19,124 gives 3.51 > 3. (Key point: Discounted payback)

## Related in Engineering Economics

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